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Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

Q8   Show that the points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.

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Given in the question,

points A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

\vec {AB }=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k

\vec {BC }=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k

\vec {CA }=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k

now let's calculate the magnitude of the vectors

|\vec {AB }|=\sqrt{4^2+2^2+6^2}=\sqrt{56}=2\sqrt{14}

|\vec {BC }|=\sqrt{6^2+3^2+9^2}=\sqrt{126}=3\sqrt{14}

|\vec {CA }|=\sqrt{10^2+5^2+15^2}=\sqrt{350}=5\sqrt{14}

As we see that AB = BC + AC, we conclude that three points are colinear.

we can also see from here,

Point B divides AC in the ratio  2 : 3.

 

 

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